UMD PDE Qualifying Exams/Jan2013PDE

Problem 1 Find the explicit solution, , of subject to . Solution Note: For notational purposes, let's put the time variable last. i.e. so that is the first variable, is the second variable. We then write our PDE as . We write the characteristic ODEs This gives Notice that this gives and which means that and must have the following form: where the coefficients are chosen so that . Also since, , then . Now, given any , we need to find such that . Clearly, we need . This means that we just need to solve the following system for Solving the second equation for gives . Substitute this into the first equation and we can solve for . We should get (after simplifying) . Therefore, . Problem 2 Let be a function. Define . a). Show that . b). Let solve for some continuous . Assume that for every , and that for . Prove that if at some , then for every . Solution a We perform a change of variables which gives: . So then differentiating and the use of Green's Formula gives: b Notation: I use to denote the average integral value symbol (dashed integral). The usual symbol used in Evans would not typeset on this wikibook. Since , . Therefore, , that is, is a supersolution to Laplace's equation. Suppose . Then by Part a, . So is a decreasing function in . Now, This estimate must hold for all . This necessarily implies since nonconstant supersolutions tend to as . Problem 3 Let solve the nonlinear eigenvalue problem Here is a 1-periodic function in all variables (that is, is the -dimensional torus) with and . a. Prove that . b. Prove that there exists no sequence of eigen-solutions such that and . Hint: Prove b by contradiction. Solution a Multiply both sides of the PDE by and integrate. . Integrate by parts to obtain: . The boundary term vanishes by the periodicity of in all variables. Thus implies that . b Assuming and our result from part a, we get This gives where the last inequality is due to Jensen's Inequality. So if , this contradicts the above inequality, i.e. we would have .